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8n^2+12n-10=0
a = 8; b = 12; c = -10;
Δ = b2-4ac
Δ = 122-4·8·(-10)
Δ = 464
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{464}=\sqrt{16*29}=\sqrt{16}*\sqrt{29}=4\sqrt{29}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{29}}{2*8}=\frac{-12-4\sqrt{29}}{16} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{29}}{2*8}=\frac{-12+4\sqrt{29}}{16} $
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